Solving Power for Solenoid w/ 9.80 mT Field

[hellomister]

Homework Statement

A solenoid 10.0 cm in diameter and 75.0 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. To produce a field of 9.80 mT at the center of the solenoid, what power must be delivered to the solenoid

I think I = 20 am I right?

Homework Equations

B=Uo * (N/l) * I

P= IV?

The Attempt at a Solution

I am stuck at getting the power, I have no clue how to get potential from a magnetic field alone?
Or if i Say P= work/time
I don't know how to get work.
And I'm not sure if my Current is correct. Any help is much appreciated.

 

[Hao]

The fact that they stated the dimensions of the copper wire implies that its properties will be relevant in the calculations.

Once the magnetic field it produced, it does not change (it is static). Hence, its energy remains constant.

This means that another source of power loss needs to be identified. With reference to the above, that source is the non-zero resistivity of Copper, which you can use to calculate the resistance of the length of Copper wire required, and hence find power with P = I^2 R

I am getting I=8.0A.
In this case, N/L is the number of turns per unit length. As the copper wires are adjacent to each other, this suggests that N/L = 1 / (diameter of copper wire).

 

[nlightner]

To solve for the power required to produce a field of 9.80 mT at the center of the solenoid, we can use the equation P = IV, where P is power, I is current, and V is potential difference. First, we need to find the current (I) in the solenoid. We can use the equation B = μ0 * (N/l) * I, where B is the magnetic field, μ0 is the permeability of free space, N is the number of turns, and l is the length of the solenoid. Rearranging this equation, we get I = B * l / (μ0 * N). Plugging in the values given, we get I = (9.80 * 10^-3 T) * (0.75 m) / (4π * 10^-7 T*m/A) * (N). Since we are given that the solenoid has a single layer of wire, we can assume that N is equal to the number of turns per unit length, which is given by N/l = π * (d/2)^2, where d is the diameter of the wire. Plugging in the values, we get N = (0.1 cm)^2 / (0.75 m) = 1.33 * 10^7 turns/m. Substituting this value into our equation for I, we get I = (9.80 * 10^-3 T) * (0.75 m) / (4π * 10^-7 T*m/A) * (1.33 * 10^7 turns/m) = 1.56 A.

Now that we have the current, we can use the equation P = IV to solve for the power. Plugging in the value for I, we get P = (1.56 A) * (V). We still need to find the potential difference (V) in order to solve for power. To do this, we can use the equation V = IR, where R is the resistance of the wire. We can find the resistance using the equation R = ρ * l / A, where ρ is the resistivity of copper (1.68 * 10^-8 Ω*m), l is the length of the wire (75 cm), and A is the cross-sectional area of the wire (π * (0.1 cm)^2